L=w+2 and 2w+2l=44
substitute equation for length in perimeter equation
2w+2(w+2)=44
2w+2w+4=44
4w+4=44
4w=40
w=10
substitute width into length equation to get length
l=10+2
l=12
a=lw
a=12*10
a=120 cm^2
The surface area of the triangular prism is: B. 72 sq. ft
<h3>Surface Area of a Triangular Prism</h3>
- The surface area of a triangular prism is given by the formula: SA = bh + (s1+s2+s3)H
- Where, b is the base, h is the height, and s1+s2+s3 is the perimeter of the triangular base, H is the length of the prism.
Thus, given the triangular prism as shown in the diagram attached below, we have the following:
b = 4 ft
h = 3 ft
H = 5 ft
s1 + s2 + s3 = 3 + 4 + 5 = 12 ft
Surface area = 4×3 + (12)5 = 12 + 60 = 72
Therefore, the surface area of the triangular prism is: B. 72 sq. ft
Learn more about the surface area of triangular prism on:
brainly.com/question/16147227
Answer:
2232.48 meters
Step-by-step explanation:
From the diagram:
Radius of semicircle = 74
Length of straightway on each side = 1000
The length of school track:
Straightway on each side = 2 * 1000 m = 2000m
Length of semicircle = πr/2
Length of both semicircle = 2 * 3.142 * 74 /2 = 232.47785
Total Length = ( 2000 + 232.47785) = 2232.48 meters
Hope this helps!
Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min